Codeforces Round #157 (Div. 2) D. Little Elephant and Elections(数位DP+枚举)

数位DP部分，不是很难。DP[i][j]前i位j个幸运数的个数。枚举写的有点搓。。。

  1 #include <cstdio>

  2 #include <cstring>

  3 using namespace std;

  4 #define LL __int64

  5 #define MOD 1000000007

  6 int dp[21][21];

  7 int p[21],num;

  8 int o[21];

  9 int dfs(int pos,int pre,int bound)

 10 {

 11     int ans,i,end;

 12     ans = 0;

 13     if(pos == -1)

 14     return pre == 0;

 15     if(!bound&&dp[pos][pre] != -1)

 16     return dp[pos][pre];

 17     end = bound ? p[pos]:9;

 18     for(i = 0;i <= end;i ++)

 19     {

 20         if(i == 4||i == 7)

 21         ans += dfs(pos-1,pre-1,bound&&i == end);

 22         else

 23         ans += dfs(pos-1,pre,bound&&i == end);

 24     }

 25     if(!bound)

 26     dp[pos][pre] = ans;

 27     return ans;

 28 }

 29 void CL(int m)

 30 {

 31     num = 0;

 32     while(m)

 33     {

 34         p[num++] = m%10;

 35         m /= 10;

 36     }

 37 }

 38 int main()

 39 {

 40     int i,m;

 41     int a1,a2,a3,a4,a5,a6,sum;

 42     LL ta1,ta2,ta3,ta4,ta5,ta6;

 43     LL ans,t1,t2;

 44     memset(dp,-1,sizeof(dp));

 45     scanf("%d",&m);

 46     CL(m);

 47     for(i = 0;i <= 9;i ++)

 48     {

 49         o[i] = dfs(num-1,i,1);

 50     }

 51     o[0] --;

 52     ans = 0;

 53     for(a1 = 0;a1 <= 9;a1 ++)

 54     {

 55         ta1 = o[a1];

 56         o[a1] -- ;

 57         for(a2 = 0;a2 <= 9&&o[a1] >= 0;a2 ++)

 58         {

 59             ta2 = o[a2];

 60             o[a2] -- ;

 61             for(a3 = 0;a3 <= 9&&o[a2] >= 0;a3 ++)

 62             {

 63                 ta3 = o[a3];

 64                 o[a3] -- ;

 65                 for(a4 = 0;a4 <= 9&&o[a3] >= 0;a4 ++)

 66                 {

 67                     ta4 = o[a4];

 68                     o[a4] -- ;

 69                     for(a5 = 0;a5 <= 9&&o[a4] >= 0;a5 ++)

 70                     {

 71                         ta5 = o[a5];

 72                         o[a5] -- ;

 73                         for(a6 = 0;a6 <= 9&&o[a5] >= 0;a6 ++)

 74                         {

 75                             ta6 = o[a6];

 76                             o[a6] -- ;

 77                             sum = a1 + a2 + a3 + a4 + a5 + a6;

 78                             if(sum <= 9&&o[a6] >= 0)

 79                             {

 80                                 t1 = ta1;

 81                                 t1 = t1*ta2%MOD;

 82                                 t1 = t1*ta3%MOD;

 83                                 t1 = t1*ta4%MOD;

 84                                 t1 = t1*ta5%MOD;

 85                                 t1 = t1*ta6%MOD;

 86                                 t2 = 0;

 87                                 for(i = sum+1;i <= 9;i ++)

 88                                 t2 += o[i];

 89                                 ans = (ans + t1*t2)%MOD;

 90                             }

 91                             o[a6] ++ ;

 92                         }

 93                         o[a5] ++ ;

 94                     }

 95                     o[a4] ++ ;

 96                 }

 97                 o[a3] ++ ;

 98             }

 99             o[a2] ++;

100         }

101         o[a1] ++ ;

102     }

103     printf("%I64d\n",ans);

104     return 0;

105 }